package Leetcode;

/*
给定一个包含非负整数的 m x n 网格，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。

        说明：每次只能向下或者向右移动一步。

        示例:

        输入:
        [
        [1,3,1],
        [1,5,1],
        [4,2,1]
        ]
        输出: 7
        解释: 因为路径 1→3→1→1→1 的总和最小。
*/

// 状态转移方程 dp[i][j] = grid[i][j] + Math.min(dp[i-1][j],dp[i][j-1]

public class problem64 {

    //时间复杂度O(outSize * inSize) 空间复杂度O(outSize * inSize)
    public static int minPathSum(int[][] grid) {
        if(grid == null || grid.length == 0){
            return 0;
        }

        int outSize = grid.length;
        int inSize = grid[0].length;

        int[][] dp = new int[outSize][inSize];
        dp[0][0] = grid[0][0];
        for(int i = 1;i < inSize;i++){
            dp[0][i] =dp[0][i-1] + grid[0][i];
        }
        for(int i = 1;i < outSize;i++){
            dp[i][0] =dp[i-1][0] + grid[i][0];
        }

        for(int i = 1;i < outSize;i++){
            for(int j = 1;j < inSize;j++){
                dp[i][j] = Math.min(dp[i-1][j],dp[i][j-1]) + grid[i][j];
            }
        }

        return dp[outSize-1][inSize-1];

    }

    public static int minPathSumCompress(int[][] grid){
        if(grid == null || grid.length == 0){
            return 0;
        }

        int outSize = grid.length;
        int inSize = grid[0].length;

        int[] dp = new int[inSize];

        for(int i = 0;i < outSize;i++){
            for(int j = 0;j < inSize;j++){
                if(i == 0 && j == 0){
                    dp[j] = grid[i][j];
                    continue;
                }
                if(i == 0){
                    dp[j] = grid[i][j] + dp[j-1];
                    continue;
                }else if(j == 0){
                    dp[j] = grid[i][j] + dp[j];
                    continue;
                }else{
                    dp[j] = grid[i][j] + Math.min(dp[j],dp[j-1]);
                    continue;
                }
            }
        }
        return dp[inSize - 1];
    }

}
